Question

Let $S_1,S_2,.....S_{101}$ be consecutive terms of an AP. If $\frac{1}{S_1{S}_2}+\frac{1}{S_2{S}_3}+\dots .+\frac{1}{S_{100}{S}_{101}}=\frac{1}{6}$and $S_1+S_{101}=50$, then $|S_1-S_{101}|$ is equal to:

• A. $10$
• B. $20$
• C. $30$
• D. $40$

Answer 1

The correct option is A

$10$

Explanation For The Correct Option:

Finding the value $|S_1-S_{101}|$

Given that $S_1,S_2,.....S_{101}$ are in AP.

Considering $d$ as common difference of AP.

$\therefore d=S_2-S_1=S_3-S_2=\dots ..=S_{101}-S_{100}$

Also given $\frac{1}{S_1{S}_2}+\frac{1}{S_2{S}_3}+\dots .+\frac{1}{S_{100}{S}_{101}}=\frac{1}{6}$

Multiplying and dividing by $d$

$$\begin{gathered} \Rightarrow \left(\frac{1}{d}\right)\left(\frac{d}{S_1{S}_2}+\frac{d}{S_2{S}_3}+\dots .+\frac{d}{S_{100}{S}_{101}}\right)=\frac{1}{6} \\ \Rightarrow \left(\frac{1}{d}\right)\left[\frac{(S_2-S_1)}{S_1{S}_2}+\frac{(S_3-S_2)}{S_2{S}_3}+\dots .+\frac{(S_{101}-S_{100})}{S_{100}{S}_{101}}\right]=\frac{1}{6} \\ \Rightarrow \left(\frac{1}{d}\right)\left[\frac{1}{S_1}-\frac{1}{S_2}+\dots .+\frac{1}{S_{100}}-\frac{1}{S_{101}}\right]=\frac{1}{6} \\ \Rightarrow \left(\frac{1}{d}\right)\left[\frac{1}{S_1}-\frac{1}{S_{101}}\right]=\frac{1}{6} \\ \Rightarrow \left(\frac{1}{d}\right)\left[\frac{S_{101}-S_1}{S_1{S}_{101}}\right]=\frac{1}{6} \\ \Rightarrow \left(\frac{1}{d}\right)\left[\frac{a+100d-a}{S_1{S}_{101}}\right]=\frac{1}{6} \\ \Rightarrow \left[\frac{100}{S_1{S}_{101}}\right]=\frac{1}{6} \\ \Rightarrow S_1{S}_{101}=600 \end{gathered}$$

$$\begin{gathered} \because |S_1–S_{101}|=\sqrt{{(S_1+S_{101})}^2-4S_1{S}_{101}} \\ =\sqrt{({50}^2-4\times 600)}\left[\because S_1+S_{101}=50\&S_1{S}_{101}=600\right] \\ =\sqrt{100} \\ =10 \end{gathered}$$

Hence, option A is the correct answer.